In the given figure, ABCD is a square, M is the midpoint of AB and `PQ_|_CM` meets AD at P and CB produced at Q. Prove that (i) PA = QB and (ii) `CP=AB+PA.`

Explanation

GIVEN A square ABCD in which M is the midpoint of `AB. PQ_|_CM` meets AD at P and CB produced at Q.
TO PROVE (i) `PA=QB and (ii)CP=AB+PA.`
PROOF (i) In `DeltaPAM and DeltaQBM,` we have
`AM=BM" "(because" M is the midpoint of AB")`
`anglePAM = angleQBM" "("each equal to "90^(@))`
`angleAMP=angleBMQ" "("vertically opp. "angles)`
`therefore" "DeltaPAM~=DeltaQBM" (AAS-criteria)."`
Hence, `PA=QB" (c.p.c.t.)."`
(ii) Join PC.
Now, in `DeltaCMP and DeltaCMQ`, we have
`PM=QM" "(because DeltaPAM~=DetlaQBM)`
`angleCMP=angleCMQ" "("each equal to "90^(@))`
`CM=CM" (common)"`
`therefore" "DeltaCMP~=DeltaCMQ" (SAS-criteria)."`
`therefore" "CP=CQ" (c.p.c.t.)."`
`rArr" "CP=CB+CQ=AB+PA" "[because CB=AB and QB=PA].`
Hence, `CP=AB+PA.`